IMO [spoiler](A)[/spoiler]
If k, m, and t are positive integers and (k/6) + (m/4) = (t/12), do t and 12 have a common factor greater than 1?
Multiply by 12 -> 2k + 3m = t
Prime factors of 12: 2,2,3
t and 12 share a prime factor other than 1?
Does t contains 2 or 3 in its prime factors?
(1) k is a multiple of 3
k -> 3,...?
2(3,...?) + 3m = t
(2,3,...?) + (3,m...?) = t
In words: Multiple of 3 + Multiple of 3 = t (THUS t must be a multiple of 3)
The rule is that a multiple of N (+/-) a multiple of N results in a multiple of N
So t -> 3,...? So YES, it shares a factor greater than 1
Sufficient
(2) m is a multiple of 3
2k + 3(3,...?) = t
(2,k,...?) + (3,3...?) = t
In words: Multiple of 2 + Multiple of 9 = t
In this case, we cannot determine if t is a multiple of 2 or 3
Example 1: 2 + 9 = 11 (neither multiple of 2, nor 3)
Example 2: 2 + 18 = 20 (multiple of 2)
Example 3: 6 + 9 = 15 (multiple of 3)
So this is insufficient
ccassel wrote:Hi,
I got the correct answer but how would you explain the solution to this question?
If k, m, and t are positve integers and k/6+m/4=t/12, do t and 12 have a common factor greater than 1?
(1) k is a multiple of 3
(2) m is a multiple of 3
Cheers,
Last edited by
MAAJ on Tue Apr 05, 2011 8:26 pm, edited 1 time in total.
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